The question is not clear, but it is possible to obtain distance, s, from the given function. This, I would show.
Answer:
s = 17 units
Step-by-step explanation:
Given f(t) = t³ - 8t² + 27t
Differentiating f(t), we have
f'(t) = 3t² - 16 t + 27
At t = 0
f'(t) = 27
This is the required obtainaible distance, s.
<span>No, because postulates are assumptions. Some true, some not. So it can't be used to prove it</span>
The vertex point is (0,4)
The two other points are (-2,0)(2,0)
Answer:
f = 2 + -15x-1 + x
Step-by-step explanation:
Simplifying
f(x) = x2 + 2x + -15
Multiply f (x)
fx = x2 + 2x + -15
Reorder the terms:
fx = -15 + 2x + x2
Solving
fx = -15 + 2x + x2
Solving for variable 'f'.
Move all terms containing f to the left, all other terms to the right.
Divide each side by 'x'.
f = -15x-1 + 2 + x
Simplifying
f = -15x-1 + 2 + x
Reorder the terms:
f = 2 + -15x-1 + x
One because the law of consines works on it
